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          <h1 class="post-title" itemprop="name headline">PAT A1012</h1>
        

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        <p>To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algebra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks – that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.</p>
<p>For example, The grades of C, M, E and A - Average of 4 students are given as the following:</p>
<p>StudentID  C  M  E  A<br>310101     98 85 88 90<br>310102     70 95 88 84<br>310103     82 87 94 88<br>310104     91 91 91 91<br>Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.</p>
<p>Input</p>
<p>Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (&lt;=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.</p>
<p>Output</p>
<p>For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.</p>
<p>The priorities of the ranking methods are ordered as A &gt; C &gt; M &gt; E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.</p>
<p>If a student is not on the grading list, simply output “N/A”.</p>
<p>Sample Input<br>5 6<br>310101 98 85 88<br>310102 70 95 88<br>310103 82 87 94<br>310104 91 91 91<br>310105 85 90 90<br>310101<br>310102<br>310103<br>310104<br>310105<br>999999<br>Sample Output<br>1 C<br>1 M<br>1 E<br>1 A<br>3 A<br>N/A<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">//#include &quot;math.h&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">using namespace std;</span><br><span class="line">struct person&#123;</span><br><span class="line">    int id;</span><br><span class="line">    int grade[4];//grade[0]:A grade[1]:C grade[2]:M grade[3]:E</span><br><span class="line">&#125;stu[2010];</span><br><span class="line"></span><br><span class="line">char course[4] = &#123;&apos;A&apos;, &apos;C&apos;, &apos;M&apos;, &apos;E&apos;&#125;;</span><br><span class="line">int Rank[1000000][4] = &#123;0&#125;;</span><br><span class="line">int now;//cmp函数中使用，表示当前按now号分数排序stu数组</span><br><span class="line"></span><br><span class="line">bool cmp(person a, person b)&#123;</span><br><span class="line">    return a.grade[now] &gt; b.grade[now];</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main()&#123;</span><br><span class="line">    int n, m;</span><br><span class="line">    scanf(&quot;%d%d&quot;, &amp;n, &amp;m);</span><br><span class="line">    //读入分数</span><br><span class="line">    for (int i = 0; i &lt; n; i++) &#123;</span><br><span class="line">        scanf(&quot;%d%d%d%d&quot;, &amp;stu[i].id, &amp;stu[i].grade[1], &amp;stu[i].grade[2], &amp;stu[i].grade[3]);</span><br><span class="line">        stu[i].grade[0] = stu[i].grade[1] + stu[i].grade[2] + stu[i].grade[3];</span><br><span class="line">    &#125;</span><br><span class="line">    for (now = 0; now &lt;=3; now++) &#123;</span><br><span class="line">        sort(stu, stu + n, cmp);//按照A C M E 分别排序</span><br><span class="line">        Rank[stu[0].id][now] = 1;</span><br><span class="line">        for (int i = 1; i &lt; n; i++) &#123;</span><br><span class="line">            //若与前一位考生分数相同</span><br><span class="line">            if(stu[i].grade[now] == stu[i - 1].grade[now])&#123;</span><br><span class="line">                Rank[stu[i].id][now] = Rank[stu[i - 1].id][now];</span><br><span class="line">            &#125;else&#123;</span><br><span class="line">                Rank[stu[i].id][now] = i + 1;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    int query;</span><br><span class="line">    for (int i = 0; i &lt; m; i++) &#123;</span><br><span class="line">        scanf(&quot;%d&quot;, &amp;query);</span><br><span class="line">        if (Rank[query][0] == 0) &#123;</span><br><span class="line">            printf(&quot;N/A\n&quot;);</span><br><span class="line">        &#125;else&#123;</span><br><span class="line">            int k = 0;</span><br><span class="line">            for (int j = 0; j &lt; 4; j++) &#123;</span><br><span class="line">                if(Rank[query][j] &lt; Rank[query][k]) k = j;</span><br><span class="line">            &#125;</span><br><span class="line">            printf(&quot;%d %c\n&quot;,Rank[query][k] ,course[k]);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<p>收获：<br>优先级为A&gt;C&gt;M&gt;E,所以考虑设置数组时就按这个顺序：<br>即0对应A,1对应C,2对应M,3对应E；<br>并且编写CMP函数的时候可以按照这个编号来排序。<br>另外，本题没有明示平均分是否需要取整以及取整方式，并且没有输出平均分，可以考虑用三门课总分代替，不用除以3。</p>

      
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